Question

by using mathematical induction answer 4^3+8^3+12^3+.....upto n terms =16n^2(n+1)^2​

Answer 1

Answer:

hope this answer will help you

Answer 2

To prove : $4^3+8^3+12^3+......................+(4n)^3 = 16n^2(n+1)^2$ is true by mathematical induction

Step-by-step explanation:

Let $P(n): 4^3+8^3+12^3+......................+(4n)^3 = 16n^2(n+1)^2$

Then let n = 1

L.H.S. is  $4^3 = 64$  R.H.S. is $16(1)^2(1+1)^2= 16\times 2^2 =16\times 4 = 64$

L.H.S.= R.H.S. Therefore P(n) is true for n = 1

Now we assume that P(n) is true for n = k

Therefore we have

$P(k): 4^3+8^3+12^3+......................+(4k)^3 = 16k^2(k+1)^2$

Now we have to prove that P(n) is true for n= k+1

$P(k+1): 4^3+8^3+12^3+......................+(4(k+1))^3 = 16(k+1)^2((k+1)+1)^2\\\\\Rightarrow P(k+1): 4^3+8^3+12^3+......................+(4(k+1))^3 = 16(k+1)^2(k+2)^2$

Taking L.H.S. we get

$4^3+8^3+12^3+......................+(4(k+1))^3\\\\\Rightarrow 4^3+8^3+12^3+......................+(4(k))^3+(4(k+1))^3$

Substituting the value of P(k) we get

$16k^2(k+1)^2 + (4(k+1))^3\\\\=16k^2(k+1)^2+4^3(k+1)^3\\\\=(k+1)^2(16k^2+64(k+1))\\\\=16(k+1)^2(k^2+4(k+1))\\\\=16(k+1)^2(k^2+4k+4)\\\\=16(k+1)^2(k+2)^2$

Which is equal to R.H.S.

Therefore by principle of mathematical induction P(n) is true for all n where n is natural number